(a) The differential equation
\frac{d^{2} y}{d t^{2}}+6 \frac{d y}{d t}+9 y=\cos t
has initial conditions, y(0)=1, y^{\prime}(0)=2. Find Y(s) and, without finding y(t), determine what functions of time will appear in the solution.
(b) If Y(s)=\frac{s+1}{s\left(s^{2}+4 s+8\right)}, find y(t).
a) Take the Laplace transform:
\left[s^{2} Y(s)-s y(0)-y^{\prime}(0)\right]+6[s Y(s)-y(0)]+9 Y(s)=\frac{s}{s^{2}+1}
\left(s^{2}+6 s+9\right) Y( s )- s (1)-2-(6)(1)=\frac{s}{s^{2}+1}
\left(s^{2}+6 s+9\right) Y(s)=\frac{s}{s^{2}+1}+s+8
\left(s^{2}+6 s+9\right) Y( s )=\frac{s+s^{3}+s+8 s^{2}+8}{s^{2}+1}
Y( s )=\frac{s^{3}+8 s^{2}+2 s+8}{(s+3)^{2}\left(s^{2}+1\right)}
To find y(t) we have to expand Y(s) into its partial fractions
Y(s)=\frac{A}{(s+3)^{2}}+\frac{B}{s+3}+\frac{C s}{s^{2}+1}+\frac{D}{s^{2}+1}
y(t)=A t e ^{-3 t}+B e ^{-3 t}+C \cos t+D \sin t
b) Y(s)=\frac{s+1}{s\left(s^{2}+4 s+8\right)}
Since \frac{4^{2}}{4}<8, there are complex factors.
\therefore complete the square in denominator
s^{2}+4 s+8=s^{2}+4 s+4+8-4
=s^{2}+4 s+4+4=(s+2)^{2}+(2)^{2} \quad\{b=2, \omega=2\}
\therefore Partial fraction expansion gives
Y(s)=\frac{A}{s}+\frac{B(s+2)}{s^{2}+4 s+8}+\frac{C}{s^{2}+4 s+8}=\frac{s+1}{s\left(s^{2}+4 s+8\right)}
Multiply by s and let s \rightarrow 0
A=1 / 8
Multiply by s\left(s^{2}+4 s+8\right)
A\left(s^{2}+4 s+8\right)+B(s+2) s+C s=s+1
A s^{2}+4 A s+8 A+B s ^{2}+2 B s +C s=s+1
s^{2}: \quad A+B=0 \quad \rightarrow \quad B=-A=-\frac{1}{8}
s^{1}: \quad 4 A+2 B+C=1 \rightarrow C=1+2\left(\frac{1}{8}\right)-4\left(\frac{1}{8}\right)=\frac{3}{4}
s^{0}: \quad 8 A=1 \quad \rightarrow A=\frac{1}{8} (This checks with above result)
Y(s)=\frac{1 / 8}{s}+\frac{(-1 / 8)(s+2)}{(s+2)^{2}+2^{2}}+\frac{3 / 4}{(s+2)^{2}+2^{2}}
y(t)=\left(\frac{1}{8}\right)-\left(\frac{1}{8}\right) e ^{-2 t} \cos 2 t+\left(\frac{3}{8}\right) e ^{-2 t} \sin 2 t